This post refers to Question 2 of the review problems at the end of Chapter 6 of George Simmon's Calculus:
Following the general form $$\int_{a}^{b} f(x)dx = \lim \limits_{max \Delta x_k\to0} \sum_{k=1}^n \Delta x_k f(x^*)$$
use Riemann sums to prove $\int_{1}^{b} \frac{1}{\sqrt{x}}dx = 2(\sqrt{b}-1)$ using equal subintervals and $x^*=\Big(\frac{\sqrt{x_{k-1}}+\sqrt{x_k}}{2}\Big)^2$.
We have $\Delta x=\frac{b-1}{n}, x_k=1+\frac{b-1}{n}k$, and as $n \to \infty$ we sum $$\sum_{k=1}^n \frac{(b-1)}{n} \frac{1}{\sqrt{x^*}} \;=\; \sum_{k=1}^n \frac{(b-1)}{n} \frac{1}{\frac{\sqrt{x_{k-1}}+\sqrt{x_k}}{2}} \;=\; \frac{2(b-1)}{n} \sum_{k=1}^n \frac{1}{\sqrt{x_{k-1}}+\sqrt{x_k}}$$
With substitution we have
$$\frac{2(b-1)}{n} \sum_{k=1}^n \frac{1}{\sqrt{{1+\frac{b-1}{n}(k-1)}}+\sqrt{1+\frac{b-1}{n}k}} \;=\; \frac{2(b-1)}{\sqrt{n}} \sum_{k=1}^n \frac{1}{\sqrt{{n+(b-1)(k-1)}}+\sqrt{n+(b-1)k}}$$
I have not been able to convert this expression into a telescoping series or anything that simplifies nicely, and would appreciate suggestions.
$\require{color}$Note $${\color{red}x_{k-1}-x_k} =\left(1+\frac{b-1}{n}(k-1)\right) -\left(1+\frac{b-1}{n}k\right) = -\frac{b-1}{n},$$ so $$\frac1{\sqrt{x_{k-1}} + \sqrt{x_k}} = \frac{\sqrt{x_{k-1}} - \sqrt{x_k}}{\color{red}x_{k-1}-x_k} = \frac{n}{b-1}\cdot\left(\sqrt{x_{k}} - \sqrt{x_{k-1}}\right).$$